Instead of saying use FOIL please, please take your time to explain this to me! PLEASE!!
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when you have the form ax^2 +bx + c, find the factors of the product ac that sum to b then factor via grouping
ac = 2 * 5 = 10
2 + 5 = 7
=> 2x^2 + 2x + 5x + 5
=> 2x(x + 1) + 5(x + 1)
=> (x + 1)(2x + 5)
ac = 2 * 5 = 10
2 + 5 = 7
=> 2x^2 + 2x + 5x + 5
=> 2x(x + 1) + 5(x + 1)
=> (x + 1)(2x + 5)
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When a quadratic has greater than unity multiplying the square use that to build a solution:
(2x + b)(x + a). 5 only has 1&5 integer factors so we must look at (2x+1)(x+5) and (2x+5)(x+1). The first expands to 2x^2+11x+5, not what we want 2x^2+7x+5.
(2x + b)(x + a). 5 only has 1&5 integer factors so we must look at (2x+1)(x+5) and (2x+5)(x+1). The first expands to 2x^2+11x+5, not what we want 2x^2+7x+5.
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(x+1)(2x+5)
see step by step solution:
http://symbolab.com/solutions/factor_qua…
see step by step solution:
http://symbolab.com/solutions/factor_qua…
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2x^2 + 7x + 5
= 2x^2 + 5x + 2x + 5
= x(2x + 5) + 1(2x + 5)
= (2x + 5)(x + 1) ANSWERT
= 2x^2 + 5x + 2x + 5
= x(2x + 5) + 1(2x + 5)
= (2x + 5)(x + 1) ANSWERT
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= 2x^2 + 2x + 5x + 5
= 2x(x + 1) + 5(x + 1)
= (x + 1)(2x + 5)
= 2x(x + 1) + 5(x + 1)
= (x + 1)(2x + 5)